Question: Complete the square to solve for $x$. $x^{2}-14x+33 = 0$
Explanation: Begin by moving the constant term to the right side of the equation. $x^2 - 14x = -33$ We complete the square by taking half of the coefficient of our $x$ term, squaring it, and adding it to both sides of the equation. Since the coefficient of our $x$ term is $-14$ , half of it would be $-7$ , and squaring it gives us ${49}$ $x^2 - 14x { + 49} = -33 { + 49}$ We can now rewrite the left side of the equation as a squared term. $( x - 7 )^2 = 16$ Take the square root of both sides. $x - 7 = \pm4$ Isolate $x$ to find the solution(s). $x = 7\pm4$ So the solutions are: $x = 11 \text{ or } x = 3$ We already found the completed square: $( x - 7 )^2 = 16$